We know the very well-known identity: $$\sum_{n=-\infty}^\infty\text{sinc}(n)=\pi.$$ But how to show that $$\sum_{n=-\infty}^\infty\text{sinc}(x+n)=\pi, \qquad \forall x?$$
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4$\begingroup$ this is indeed a remarkable result: no matter how much you shift your sample points on a sinc function, the sum of those samples is constant. For a derivation, see math.stackexchange.com/questions/1242280/… $\endgroup$– Carlo BeenakkerCommented Mar 5, 2019 at 17:41
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